Master AP Chemistry Topic 7.7 with this free practice quiz. Test your skills on ice tables, perfect squares, small x approximation, and solving for equilibrium values to prepare for your AP exam.
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The ICE Table Turns K into One Equation in x
Given K and a starting mixture, the equilibrium concentrations are the unknowns. The ICE table organises the bookkeeping in three rows: Initial (the starting concentrations), Change (written in terms of a single unknown, x), and Equilibrium (the initial row plus the change row).
Only the equilibrium row goes into the K expression. Substituting it gives one equation in one unknown; the rest of this topic is choosing a legal way to solve it.
The Change Row Carries the Coefficients
The variable x is the change in concentration tied to the extent of reaction. Each species changes by its own coefficient times x: reactants fall, products rise, and the mole ratio is preserved exactly.
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) starting with [N₂] = 0.40, [H₂] = 1.20 and no NH₃, the equilibrium row is [N₂] = 0.40 − x, [H₂] = 1.20 − 3x, [NH₃] = 2x. If H₂ fell by only x, the row would destroy hydrogen atoms: each x of reaction consumes 3x of H₂ per x of N₂.
Products present at the start change nothing: they appear in the initial row, and their change entries still follow their coefficients.
Perfect Squares: Take the Square Root of Both Sides
Some expressions are a ratio of squares. When the equilibrium row produces (2x)² over (a − x)², taking the square root of both sides turns the equation into a linear one; no quadratic is needed.
Watch for the pattern rather than the reaction: squared terms top and bottom. The square root step also produces a second, negative candidate value; it is tested like any other root.
The Quadratic Formula, and Rejecting Roots
When the expression is not a perfect square, expand it into the standard form ax² + bx + c = 0 and use the quadratic formula. Two roots come out, and algebra alone cannot choose between them.
A root is physical only if every equilibrium concentration stays positive and no species holds more atoms than the starting mixture supplied. A root that makes a concentration negative is discarded, however clean it looks. If both roots fail, the error is in the ICE table, usually a sign or a missing coefficient in the change row.
The Small-x Approximation and the 5 Percent Check
When K is very small, the reaction barely moves and x is tiny. A term like (0.50 − x) can then be replaced by 0.50, and the equation collapses to something solvable by a square root.
The check that follows is not optional: x must be less than 5 percent of the concentration it was subtracted from. If it is, the neglected amount was genuinely negligible. If it is not, the quadratic must be solved instead. The approximation can even return an x larger than the starting concentration, which is impossible: no reactant can lose more than it had.
Check the Answer by Substituting Back
The finished equilibrium row, substituted into the K expression, must return the stated K within rounding. This check catches wrong roots, dropped coefficients and misread squares, because all of them disturb the ratio.
Two further checks cost nothing: no equilibrium concentration may be negative, and the atoms in the equilibrium row must match the atoms in the initial row. For a very large K the row shows reactants nearly, but never exactly, used up.
Worked Examples
Example 1: a perfect square. For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 64.0, starting with [H₂] = [I₂] = 0.50 mol L⁻¹ and no HI. The equilibrium row is 0.50 − x, 0.50 − x, 2x, so Kc = (2x)² / (0.50 − x)² = 64. Taking square roots: 2x / (0.50 − x) = 8, so 2x = 4.0 − 8x and x = 0.40. Then [HI] = 0.80 and [H₂] = [I₂] = 0.10 mol L⁻¹. Check: (0.80)² / (0.10 × 0.10) = 64.
Example 2: a quadratic, and a failed 5 percent check. For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 0.042, starting with [PCl₅] = 1.00 mol L⁻¹. The small-x estimate is x = √0.042 = 0.205 M, which is 20.5 percent of 1.00 M, so the approximation fails. The full equation x² / (1.00 − x) = 0.042 rearranges to x² + 0.042x − 0.042 = 0, whose positive root is x = 0.185 M. Then [PCl₅] = 0.815 and [PCl₃] = [Cl₂] = 0.185 mol L⁻¹, an 18.5 percent dissociation. Check: (0.185)² / 0.815 = 0.042.
Example 3: a valid small-x approximation. For SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g), Kc = 2.0 × 10⁻⁵, starting with [SO₂Cl₂] = 0.50 mol L⁻¹. Neglecting x next to 0.50 gives x² / 0.50 = 2.0 × 10⁻⁵, so x = √(2.0 × 10⁻⁵ × 0.50) = 3.2 × 10⁻³. The check: 3.2 × 10⁻³ is 0.64 percent of 0.50, well under 5 percent, so the approximation stands and [Cl₂] = [SO₂] = 3.2 × 10⁻³ mol L⁻¹. The equilibrium [SO₂Cl₂] is 0.50 − 0.0032 = 0.497 mol L⁻¹.
Example 4: both roots positive, one impossible. For 2NO₂(g) ⇌ N₂O₄(g), Kc = 25, starting with [NO₂] = 0.60 mol L⁻¹ and no N₂O₄. The row is [NO₂] = 0.60 − 2x, [N₂O₄] = x, giving x / (0.60 − 2x)² = 25, or 100x² − 61x + 9 = 0. The roots are x = 0.25 and x = 0.36, both positive. But x = 0.36 sets [NO₂] = 0.60 − 0.72 = −0.12 mol L⁻¹, so it is rejected: it consumes more NO₂ than the vessel holds. With x = 0.25, [NO₂] = 0.10 and [N₂O₄] = 0.25 mol L⁻¹. Check: 0.25 / (0.10)² = 25.
Traps That Cost Marks in Topic 7.7
“Every species changes by x.” Species change by their coefficient times x. In the ammonia synthesis the H₂ change is 3x, not x, and writing x breaks the atom balance of the whole table.
“Substitute the initial row into K.” K is built from equilibrium concentrations only. The initial row gives Q at the first instant, and if a product starts absent that value is 0: a statement about direction, never a value of K.
“Neglect x whenever K looks small.” The licence to neglect comes from the 5 percent check run after solving, not from the look of K. For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) with Kc = 0.042 and [PCl₅] = 1.00 mol L⁻¹, neglecting x gives 20.5 percent dissociation; the check fails, and the quadratic gives 18.5 percent.
“Keep the smaller root, or the tidier root.” Roots are chosen by the equilibrium row, not by size or neatness. Keep the root that leaves every concentration positive and reachable; if both fail, rebuild the ICE table.
“(2x)² can be written as 2x².” The whole concentration is squared: (2x)² = 4x². Writing 2x² understates the product term by a factor of 2 and shifts the solved x with it.
“A huge K means the reaction goes to completion, so no solving is needed.” Equilibrium still leaves a trace of reactant. For H₂(g) + Br₂(g) ⇌ 2HBr(g) with Kc = 3.8 × 10⁴ and both reactants at 0.10 mol L⁻¹, the square root method gives [HBr] = 0.198 mol L⁻¹ with 0.0010 mol L⁻¹ of each reactant left, not 0.20 and zero.
Key Relationships
- ICE rows: equilibrium = initial + change, and only the equilibrium row enters the K expression
- Every change entry is the species coefficient times x, negative for reactants and positive for products in the direction written
- A ratio of squares in the K expression is solved by taking the square root of both sides, giving a linear equation
- A non-square expression is expanded to ax² + bx + c = 0 and solved with the quadratic formula
- A root is accepted only when every equilibrium concentration it produces is positive and atom-balanced against the initial row
- The small-x approximation replaces (c − x) with c, valid only when x is under 5 percent of c
- A failed 5 percent check, or an approximate x larger than the starting concentration, sends the problem back to the quadratic
- The solved equilibrium row must return the stated K on back-substitution; that check closes every problem in this topic